Question

Brainliest to whoever answers correctly.
A buoy is constructed out of the bottom half of a sphere with
a cone on top. The radius of the sphere and the radius of the
cone is 9 ft. The height of the buoy is 15 ft.
What is the volume of the buoy?
Enter your answer, in exact form, in the box.
ft3

Answer 1

• r=9
• h=15

Volume of semisphere

$\\ \rm\dashrightarrow \dfrac{2}{3}\pi r^3$

$\\ \rm\dashrightarrow \dfrac{2}{3}\pi (9)^3$

$\\ \rm\dashrightarrow \dfrac{2}{3}\pi 729$

$\\ \rm\dashrightarrow 2(243)\pi$

$\\ \rm\dashrightarrow 486\pi ft^3$

Volume of cone

$\\ \rm\dashrightarrow \dfrac{1}{3}\pi r^2h$

$\\ \rm\dashrightarrow \dfrac{1}{3}\pi (9)^2(15)$

$\\ \rm\dashrightarrow {27(15)\pi}$

$\\ \rm\dashrightarrow 405\pi ft^3$

Total volume

$\\ \rm\dashrightarrow (486+405)\pi$

$\\ \rm\dashrightarrow 891\pi$

$\\ \rm\dashrightarrow 2797.74ft^3$

Answer 2

Answer:

$\sf V=891\pi \:\:ft^3$

Step-by-step explanation:

Formulas

$\textsf{Volume of a cone}=\sf \dfrac{1}{3} \pi r^2 h$

$\textsf{Volume of a hemisphere}=\sf \dfrac{2}{3} \pi r^3$

(where r is the radius and h is the height)

Given:

• r = 9 ft
• h = 15 ft

$\begin{aligned}\textsf{Volume of buoy} & =\textsf{Volume of cone}+\textsf{Volume of hemisphere}\\\\\implies \sf V & = \sf \dfrac{1}{3} \pi r^2 h + \dfrac{2}{3} \pi r^3\\\\& = \sf \dfrac{1}{3} \pi (9)^2 (15) + \dfrac{2}{3} \pi (9)^3\\\\& = \sf 405 \pi + 486 \pi\\\\& = \sf 891\pi \:\:ft^3\end{aligned}$